document.write( "Question 842109: Find the equation of the tangent to the curve y = e^x at the point where it crosses the Y-axis.\r
\n" ); document.write( "\n" ); document.write( "I differentiated y = e^x (which is e^x), which gives you your gradient for the tangent..\r
\n" ); document.write( "\n" ); document.write( "Subbing in x= 0 into the original equation;
\n" ); document.write( "y = e^(0) = 1
\n" ); document.write( "Therefore if y=1, x=0 (exponential function)\r
\n" ); document.write( "\n" ); document.write( "Sub into the formula:
\n" ); document.write( "y-y1 = m(x-x1)
\n" ); document.write( "y-1 = e^x(x-0)\r
\n" ); document.write( "\n" ); document.write( "y= xe^x + 1\r
\n" ); document.write( "\n" ); document.write( "However the answers say y = x+1. Whilst in theory, i completely understand whyh it would be x+1, i can't seem to resolve the question quantatively. Thank you for your help! \n" ); document.write( "

Algebra.Com's Answer #507367 by ewatrrr(24785) $\"\"$ $\"About$

You can put this solution on YOUR website!

 
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document.write( "Hi,
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document.write( "\"I differentiated y = e^x (which is e^x), which gives you your gradient for the tangent\"(line)
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document.write( " = e^x and when x = 0 ⇒  = 1
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document.write( "The \"tangent to the curve is a LINE:  y-1 = (x-0)  0r y = x + 1
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